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Magazine Chapter 10: Problem 11
If the function \(f(t)\) is a solution of the initial-value problem \(y^{\prime}=2 y-3, y(0)=4,\) find \(f(0)\) and \(f^{\prime}(0)\).
Short Answer
Expert verified
The value of \(f(0)\) is 4 and \(f'(0)\) is 5.
Step by step solution
01
- Identify the type of differential equation
Recognize that the given initial-value problem is a first-order linear differential equation: \(y' = 2y - 3\) with an initial condition \(y(0) = 4\).
02
- Solve the homogeneous equation
First, find the complementary solution to the homogeneous equation \(y' = 2y\). The general solution to this is \(y_h = Ce^{2t}\), where \(C\) is a constant.
03
- Find a particular solution
Find a particular solution to the non-homogeneous equation. Since the non-homogeneous term is a constant, assume a particular solution of the form \(y_p = A\). Substituting into the differential equation: \(0 = 2A - 3 \rightarrow A = \frac{3}{2}\).So, the particular solution is \(y_p = \frac{3}{2}\).
04
- Form the general solution
Combine the complementary and particular solutions to get the general solution: \(y(t) = y_h + y_p = Ce^{2t} + \frac{3}{2}\).
05
- Apply the initial condition
Use the initial condition \(y(0) = 4\) to find \(C\). Substitute \(t = 0\): \(4 = Ce^0 + \frac{3}{2} \rightarrow 4 = C + \frac{3}{2} \rightarrow C = \frac{5}{2}\).Thus, the solution is \(y(t) = \frac{5}{2}e^{2t} + \frac{3}{2}\).
06
- Find \(f(0)\)
To find \(f(0)\), substitute \(t = 0\) in the solution: \(f(0) = y(0) = \frac{5}{2}e^0 + \frac{3}{2} = \frac{5}{2} + \frac{3}{2} = 4\).
07
- Find \(f'(0)\)
Differentiate the solution \(y(t) = \frac{5}{2} e^{2t} + \frac{3}{2}\). \(y'(t) = \frac{5}{2} \times 2e^{2t} = 5 e^{2t}\). Substitute \(t = 0\): \(f'(0) = y'(0) = 5 e^0 = 5\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-order linear differential equation
When tackling the problem, we first identify that it involves a first-order linear differential equation. These are equations of the form \( y' + p(t)y = q(t) \). In this case, we have \( y' = 2y - 3 \). Notice that it is first-order because the highest derivative is \( y' \) and it is linear as it fits the general form.
Solving these equations typically involves finding the general solution to the homogeneous part and then adding a particular solution to account for the non-homogeneous part.
We should also note that the equation has an initial condition: \( y(0) = 4 \). This will help us determine the constant in our general solution after solving the equation.
Solving these equations typically involves finding the general solution to the homogeneous part and then adding a particular solution to account for the non-homogeneous part.
We should also note that the equation has an initial condition: \( y(0) = 4 \). This will help us determine the constant in our general solution after solving the equation.
Particular solution
The particular solution addresses the non-homogeneous aspect of our differential equation. To find it, we look at the non-homogeneous term, which in this instance is \(-3\).
Since this term is a constant, we guess a particular solution of the form \( y_p = A \), where \( A \) is a constant we'll solve for. By substituting \( y_p \) into the differential equation, we derive that \( A = \frac{3}{2} \).
This gives us our particular solution: \( y_p = \frac{3}{2} \). In other words, this is a specific solution to the differential equation that also factors in the continuous term \( -3 \).
Since this term is a constant, we guess a particular solution of the form \( y_p = A \), where \( A \) is a constant we'll solve for. By substituting \( y_p \) into the differential equation, we derive that \( A = \frac{3}{2} \).
This gives us our particular solution: \( y_p = \frac{3}{2} \). In other words, this is a specific solution to the differential equation that also factors in the continuous term \( -3 \).
Homogeneous equation
Next, we solve the homogeneous part of the differential equation, which is \( y' = 2y \). This kind of problem assumes \( q(t) = 0 \).
To solve it, we usually assume a solution of the form \( y_h = Ce^{kt} \). For our current equation, \( k = 2 \), leading to \( y_h = Ce^{2t} \) where \( C \) is an arbitrary constant.
This complementary or homogeneous solution tackles the 'pure' equation void of any external forcing terms. Combining this with the particular solution provides us with a full general solution.
To solve it, we usually assume a solution of the form \( y_h = Ce^{kt} \). For our current equation, \( k = 2 \), leading to \( y_h = Ce^{2t} \) where \( C \) is an arbitrary constant.
This complementary or homogeneous solution tackles the 'pure' equation void of any external forcing terms. Combining this with the particular solution provides us with a full general solution.
Initial condition
Finally, the initial condition is crucial for determining the constant \( C \) in our general solution. Our initial condition is \( y(0) = 4 \).
We substitute \( t = 0 \) into our general solution \( y(t) = Ce^{2t} + \frac{3}{2} \):
\( 4 = Ce^0 + \frac{3}{2} \)
Solving this, we find \( C = \frac{5}{2} \).
This step ensures that the solution doesn't just fit the equation generally but also satisfies the specific initial requirement.
Our final specific solution becomes \( y(t) = \frac{5}{2}e^{2t} + \frac{3}{2} \), fully integrating the equation's dynamics and the given starting condition.
We substitute \( t = 0 \) into our general solution \( y(t) = Ce^{2t} + \frac{3}{2} \):
\( 4 = Ce^0 + \frac{3}{2} \)
Solving this, we find \( C = \frac{5}{2} \).
This step ensures that the solution doesn't just fit the equation generally but also satisfies the specific initial requirement.
Our final specific solution becomes \( y(t) = \frac{5}{2}e^{2t} + \frac{3}{2} \), fully integrating the equation's dynamics and the given starting condition.
